VIJOSP1471 's Playground

VIJOSP1471 's Playground

Links: https://vijos.org/p/1471

Ideas

Recursion.

First find the leftmost one can jump to the rear of the L,

Then the points after the L have two cases: either a enough to jump to the rear steps of 1, or can be a pace to l have L jump to the rear of the number of steps to 2.

For points prior to L, the same operation would be equivalent to reducing R to L in the code.

Code

1#include <iostream>2#include <cstdio>3 using namespacestd;4 5 Const intMAXN =100000+Ten;6 7 intn,q;8 intA[MAXN],STEP[MAXN];9 TenInlineintRead_int () { One     CharC=GetChar (); A      while(!isdigit (c)) c=GetChar (); -     intx=0; -      while(IsDigit (c)) { thex=x*Ten+c-'0'; -C=GetChar (); -     } -     returnx; + } -  + intMain () { AIos::sync_with_stdio (false); atN=read_int (); q=read_int (); -      for(intI=1; i<=n;i++) a[i]=read_int (); -      -     intL=1, r=n+1; -step[r]=0; -      Do{ in         intI=1; -          while(I+a[i]<r) i++; to          for(intj=i;j<r;j++) +            if(j+a[j]>=r) step[j]=step[r]+1; -            Elsestep[j]=step[r]+2; theR=i; *} while(r>1); $     intx;Panax Notoginseng      for(intI=1; i<=q;i++) { -x=read_int (); thecout<<Step[x]; +         if(I&LT;Q) cout<<" "; A         Else  thecout<<"\ n"; +     } -     return 0; $}

VIJOSP1471 's Playground